Question

If the nth term of the AP 9, 7, 5….. is same as the nth term of the AP 15, 12, 9,…., find n.
(a) . 7
(b). 8
(c). 9
(d). 10

Answer 1

Hint: In this question use the formula of finding the \[{{n}^{th}}\] term by \[{{T}_{n}}=a+\left( n-1 \right)d\], where a is \[{{1}^{st}}\] term, d is common difference and n is number of terms for both the series and then equate it to get the value of n.

Complete step-by-step answer:

In the question we are given two arithmetic series 9, 7, 5…… and 15, 12, 9…… We are further said that for a particular value of n, the \[{{n}^{th}}\] term of both the series will be equal.
At first we briefly understood what is arithmetic progression.
In mathematics, an arithmetic progression (AP) or an arithmetic sequence of numbers such that the difference between the consecutive terms is constant. Here difference means the second minus first. For instance, the sequence 5, 7, 9, 11, 13, 15…… is an arithmetic progression with a common difference of 2.
If the initial term of an arithmetic is considered as a and common difference in successive terms is d, then the \[{{n}^{th}}\] term of a sequence (\[{{T}_{n}}\]) is denoted by:
\[{{T}_{n}}=a+\left( n-1 \right)d\] and in general, \[{{T}_{n}}={{T}_{m}}+\left( n-m \right)d\].
A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes called an arithmetic progression. The sum of finite arithmetic progression is called an arithmetic series.
Here for the question we will find the \[{{n}^{th}}\] term using the formula, \[{{T}_{n}}=a+\left( n-1 \right)d\], where a is the first term of series, n is the number of terms and d is common difference.
Now for the series 9, 7, 5…. The first term is 9 and the common difference is (7 - 9) or -2. So it’s \[{{n}^{th}}\] term will be,
\[{{T}_{n}}=9+\left( n-1 \right)\left( -2 \right)\], which we got by using formula of finding \[{{n}^{th}}\] term \[{{T}_{n}}=a+\left( n-1 \right)d\].
So, the value of \[{{n}^{th}}\] term is,
\[\Rightarrow {{T}_{n}}=9-2n+2\]
Or, \[{{T}_{n}}=11-2n\]
Hence, the \[{{n}^{th}}\] term for the series 9, 7, 5…… is (11 – 2n).
Now for the series 15, 12, 9…… the first term is 15 and the common difference is (12 - 15) or -3. So it’s \[{{n}^{th}}\] term will be,
 \[\Rightarrow {{T}_{n}}=15+\left( n-1 \right)\left( -3 \right)\]
Which we got by using the formula of finding \[{{n}^{th}}\] term, \[{{T}_{n}}=a+\left( n-1 \right)d\].
So, the value of \[{{n}^{th}}\] term is,
\[\Rightarrow {{T}_{n}}=15-3n+3\]
Or, \[{{T}_{n}}=18-3n\]
Hence the \[{{n}^{th}}\] term for the series,
15, 12, 9…… is (18 – 3n).
Now we will equate both the \[{{n}^{th}}\] terms to solve for the value of n so we get,
\[\Rightarrow 11-2n=18-3n\]
Now we will add 3n to both sides so we get,
\[\Rightarrow 11-2n+3n=18-3n+3n\]
Or, \[11+n=18\].
So, the value of n is (18 - 11) or 7.
Hence the correct option is (a).

Note: The behavior of arithmetic progression depends on common difference d. If common difference is positive then terms will go towards infinity and if the common difference is negative then terms will go towards negative infinity.